Question

A solid cylinder rolls up an inclined plane of inclination $\theta$with an initial velocity v. How far does the cylinder go up the plane?

• A.
• B. $\frac { v ^ { 2 } } { 4 g \sin \theta }$
• C. $\frac { 3 v ^ { 2 } } { g \sin \theta }$
• D. $\frac { 3 \mathrm { v } ^ { 2 } } { 4 \mathrm {~g} \sin \theta }$

Answer 1

Answer: (D)

$\frac { 3 \mathrm { v } ^ { 2 } } { 4 \mathrm {~g} \sin \theta }$

Answer 2

Let the cylinder go up the plane upto a height h let M and R be the mass and radius of the cylinder respectively. According to law of conservation of mechanical energy we get

$\frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } + \frac { 1 } { 2 } \mathrm { I } \omega ^ { 2 } = \mathrm { Mgh }$

$\frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } + \frac { 1 } { 2 } \frac { \mathrm { MR } ^ { 2 } } { 2 } \omega ^ { 2 } = \mathrm { Mgh }$ $\frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } + \frac { 1 } { 4 } \mathrm { MR } ^ { 2 } \omega ^ { 2 } = \mathrm { Mgh }$

$\frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } + \frac { 1 } { 4 } \mathrm { Mv } ^ { 2 } = \mathrm { Mgh }$ $( \because \mathrm { v } = \mathrm { R } \omega )$ $\frac { 3 } { 4 } \mathrm { Mv } ^ { 2 } = \mathrm { Mgh }$

$h = \frac { 3 v ^ { 2 } } { 4 g }$ …(i)

Let s be distance travelled by the cylinder up the plane then

$\sin \theta = \frac { h } { s }$ or $s = \frac { h } { \sin \theta } = \frac { 3 v ^ { 2 } } { 4 g \sin \theta }$ (Using(i))